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If the energy of a single photon is given by $E=hf$ then a linearly polarized photon has an energy E which is dependent on only its frequency.

But a circularly polarized photon also has this same energy E due to its frequency, but has "extra energy" because it is rotating.

I know there must be a flaw in my logic somewhere, because this is saying that two photons with the same frequency can have different energies depending on their polarization.

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    $\begingroup$ The photon isn't "rotating", and it doesn't have "extra energy". You didn't actually present a logic we could point out a flaw in, because you just asserted the photon has "extra energy" without deriving that from anything. $\endgroup$
    – ACuriousMind
    Aug 25, 2016 at 19:21
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    $\begingroup$ Sounds plausible to me. Fair question. $\endgroup$ Aug 25, 2016 at 19:44
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    $\begingroup$ Classically, the energy density of the radiation field is independent of the nature of its polarization, which provides a less mathematically demanding way to understand the system. $\endgroup$ Aug 25, 2016 at 20:33
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    $\begingroup$ @BobBee Your claim seems to contradict dmckee's. His seems that photons are intrinsically circularly polarized (have helicity). That would mean circular polarization consists of pure state photons, while linear polarization would be the linear combination you're suggesting. $\endgroup$
    – Real
    Aug 26, 2016 at 4:46
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    $\begingroup$ @Real Both are true, no contradiction. In QM you can choose the single helicity states or linearly polarized states as the basis of the Hilbert space. One is just a superposition of the other. The difference is their states -- a linearly polarized photon is a superposition of two circularly polarized photons. Both can be just a single photon. In classical EM it's similar, either can be the basis for an expansion of the wave as a sum of the other. See it also in the answer at physics.stackexchange.com/questions/107303/…. $\endgroup$
    – Bob Bee
    Aug 26, 2016 at 16:15

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All photons have the same energy which depends only on the frequency E = h·f. The polarization of a photon has no influence on the energy even though it is related to its spin. So a linearly polarized photon has the same energy as a left or right circularly polarized photon which corresponds to the spin being parallel or antiparallel to the propagation direction.

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